ÌâÄ¿ÄÚÈÝ

µª»¯ÂÁ(AlN)¾ßÓÐÄ͸ßΡ¢¿¹³å»÷¡¢µ¼ÈÈÐԺõÈÓÅÁ¼ÐÔÖÊ£¬±»¹ã·ºÓ¦ÓÃÓÚµç×Ó¹¤Òµ¡¢Ìմɹ¤ÒµµÈÁìÓò¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬µª»¯ÂÁ¿Éͨ¹ýÈçÏ·´Ó¦ºÏ³É£ºAl2O3£«N2£«3C=2AlN£«3CO£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)¡£

A£®ÔÚµª»¯ÂÁµÄºÏ³É·´Ó¦ÖУ¬N2ÊÇ»¹Ô­¼Á£¬Al2O3ÊÇÑõ»¯¼Á
B£®ÉÏÊö·´Ó¦ÖÐÿÉú³É2 mol AlN£¬N2µÃµ½3 molµç×Ó
C£®µª»¯ÂÁÖеªÔªËصĻ¯ºÏ¼ÛΪ£­3¼Û
D£®µª»¯ÂÁÊôÓÚ¸´ºÏ²ÄÁÏ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÔÚÔªËØÖÜÆÚ±íÖУ¬ÂÁÔÚÅðµÄÕýÏ·½£¬ËüÃÇÓкܶàÏàËƵĻ¯Ñ§ÐÔÖÊ£¬¿ÉÒÔÐγÉÐí¶à×é³ÉºÍÐÔÖÊÀàËƵĻ¯ºÏÎï¡£µ¥ÖÊÅð¿ÉÒÔͨ¹ýÅðþ¿óMg2B2O5¡¤H2OÀ´ÖÆÈ¡¡£

£¨1£©Ð´³ö²½Öè¢ÙµÄ»¯Ñ§·½³Ìʽ                         £¬
£¨2£©Ð´³ö²½Öè¢ÚµÄÀë×Ó·½³Ìʽ                         £¬
£¨3£©Ð´³ö±íʾÈÜÒºaÎïÁÏÊغãµÄ¹Øϵ                    £»
£¨4£©²½Öè¢ÛÖмÓÈëÏ¡H2SO4µÄ×÷ÓÃÊÇ                   £»
£¨5£©½«ÖƵõĴÖÅðÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦È«²¿Éú³ÉBI3£¬BI3ÈÈ·Ö½â¿ÉÒԵõ½´¿¾»µÄµ¥ÖÊÅð¡£0£®25 g´ÖÅðÖƳɵÄBI3·Ö½âµÃµ½µÄI2È«²¿±»ÊÕ¼¯ºó£¬ÓÃ2.00 mol/L Na2S2O3ÈÜÒºµÎ¶¨£¬´ïµ½µÎ¶¨ÖÕµãʱÏûºÄ27.00 mL Na2S2O3ÈÜÒº¡££¨ÒÑÖª£ºI2+2S2O32£­2I£­+S4O62£­£©
¢ÙµÎ¶¨¹ý³ÌÖÐËùÓõÄָʾ¼ÁΪ         £¬µÎ¶¨ÖÕµãµÄÏÖÏóÊÇ                £»
¢Ú´ÖÅðÖÐÅðµÄº¬Á¿Îª____          £»
£¨6£©ÀûÓÃÅðÉ°¾§Ì壨Na2B4O7?10H2O£©¿ÉÖƱ¸¹ýÅðËáÄÆ£¬ËüÊÇÒ»ÖÖÓÅÁ¼µÄƯ°×¼Á£¬±»¹ã·ºÓ¦ÓÃÓÚÏ´Ò·ۡ¢Æ¯°×·Û¡¢Ï´µÓ¼ÁÖС£ÒÑÖª´¿¾»µÄ¹ýÅðËáÄƾ§ÌåÖи÷ÔªËصÄÎïÖʵÄÁ¿Ö®±ÈΪÒÔn£¨Na£©£ºn£¨B£©£ºn£¨H£©£ºn£¨O£©=1£º1£ºn£º7¡£È¡¹ýÅðËáÄƾ§ÌåÔÚ70¡æÒÔÉϼÓÈȽ«Öð²½Ê§È¥½á¾§Ë®£¬²âµÃ¹ÌÌåÖÊÁ¿Ëæζȵı仯ÈçͼËùʾ£¬ÔòÄËʱËùµÃ¾§ÌåµÄ»¯Ñ§Ê½Îª             £¬Çëд³ö¼òÒªµÄ½âÌâ¹ý³Ì¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø